[proofplan] A Dedekind domain is a (i) Noetherian, (ii) integrally closed, (iii) integral domain in which every nonzero prime ideal is maximal. We verify each condition for $\mathcal{O}_L$. Noetherianity follows from $\mathcal{O}_L \cong \mathbb{Z}^n$ as an abelian group (by [Existence of an Integral Basis](/theorems/1581)): every ideal is a subgroup of $\mathbb{Z}^n$, hence finitely generated as an abelian group and therefore as an ideal. Integral closure follows from transitivity of integral dependence over $\mathbb{Z}$: an element of $L$ integral over $\mathcal{O}_L$ is integral over $\mathbb{Z}$, hence lies in $\mathcal{O}_L$ by definition. The "every nonzero prime is maximal" condition reduces to the fact that nonzero quotients $\mathcal{O}_L/\mathfrak{p}$ are finite (by [Nonzero Ideals Have Bounded Quotients](/theorems/1583)), because a finite integral domain is a field. [/proofplan] [step:State the Dedekind domain axioms and collect what must be shown] A commutative ring $R$ is a **Dedekind domain** if: \begin{align*} &(i)\;\; R \text{ is an integral domain;} \\ &(ii)\;\; R \text{ is Noetherian;} \\ &(iii)\;\; R \text{ is integrally closed in its fraction field;} \\ &(iv)\;\; \text{every nonzero prime ideal of } R \text{ is maximal.} \end{align*} We prove each condition holds for $R = \mathcal{O}_L$, noting the fraction field is $\operatorname{Frac}(\mathcal{O}_L) = L$ by [Fraction Field of the Ring of Integers](/theorems/1571). [guided] We work through the four axioms in turn. Let us set up scope and notation. - $L/\mathbb{Q}$ is a number field of degree $n = [L : \mathbb{Q}]$. - $\mathcal{O}_L \subseteq L$ is its ring of integers: the set of $\alpha \in L$ that are algebraic integers over $\mathbb{Z}$. - By [Ring of Integers is a Ring](/theorems/1567), $\mathcal{O}_L$ is a commutative ring (with $1 \in \mathcal{O}_L$). - By [Fraction Field of the Ring of Integers](/theorems/1571), $\operatorname{Frac}(\mathcal{O}_L) = L$. - By [Existence of an Integral Basis](/theorems/1581), $\mathcal{O}_L \cong \mathbb{Z}^n$ as abelian groups — there exists an integral basis $\omega_1, \ldots, \omega_n \in \mathcal{O}_L$ with every element of $\mathcal{O}_L$ uniquely expressible as a $\mathbb{Z}$-combination of them. These four ingredients, plus the deeper finiteness-of-quotients lemma stated in [Nonzero Ideals Have Bounded Quotients](/theorems/1583), are all the tools we need. [/guided] [/step] [step:Verify (i): $\mathcal{O}_L$ is an integral domain] By [Ring of Integers is a Ring](/theorems/1567), $\mathcal{O}_L$ is a commutative subring of $L$ containing $1$. As a subring of the field $L$, it inherits the property of having no zero divisors: if $\alpha, \beta \in \mathcal{O}_L \subseteq L$ with $\alpha\beta = 0$, then in $L$ (a field, hence an integral domain) either $\alpha = 0$ or $\beta = 0$. Moreover $1 \neq 0$ in $L$, hence in $\mathcal{O}_L$. So $\mathcal{O}_L$ is an integral domain. [guided] Axiom (i) asks that $\mathcal{O}_L$ be a nonzero commutative ring with no zero divisors. *Commutative ring with $1$*: confirmed by [Ring of Integers is a Ring](/theorems/1567), which shows $\mathcal{O}_L$ is a subring of $L$ containing $1$ and closed under sums, differences, and products. *No zero divisors*: a subring of an integral domain has no zero divisors. The argument is immediate: $L$ is a field — in particular has no zero divisors — and if $\alpha\beta = 0$ holds inside $\mathcal{O}_L$ it also holds inside $L$, so $\alpha = 0$ or $\beta = 0$ in $L$, hence in $\mathcal{O}_L$ (since $\mathcal{O}_L \subseteq L$). *Nonzero*: $1 \in \mathcal{O}_L$ is distinct from $0$ in $L$, hence in $\mathcal{O}_L$. So (i) holds. [/guided] [/step] [step:Verify (ii): $\mathcal{O}_L$ is Noetherian] A ring is Noetherian iff every ideal is finitely generated. Let $\mathfrak{a} \unlhd \mathcal{O}_L$ be any ideal. By [Existence of an Integral Basis](/theorems/1581), there is an isomorphism of abelian groups \begin{align*} \Phi: \mathbb{Z}^n \xrightarrow{\sim} \mathcal{O}_L. \end{align*} The preimage $\Phi^{-1}(\mathfrak{a})$ is a subgroup of $\mathbb{Z}^n$. By the [structure theorem for finitely generated abelian groups](/theorems/???) (or more directly: subgroups of $\mathbb{Z}^n$ are free of rank $\leq n$), every subgroup of $\mathbb{Z}^n$ is itself a finitely generated abelian group. Hence $\mathfrak{a}$ is finitely generated as an abelian group, say $\mathfrak{a} = \mathbb{Z} x_1 + \cdots + \mathbb{Z} x_m$ with $x_1, \ldots, x_m \in \mathfrak{a}$. A generating set as an abelian group is a fortiori a generating set as an ideal: since $\mathbb{Z} \subseteq \mathcal{O}_L$, \begin{align*} \mathfrak{a} = \mathbb{Z} x_1 + \cdots + \mathbb{Z} x_m \subseteq \mathcal{O}_L x_1 + \cdots + \mathcal{O}_L x_m \subseteq \mathfrak{a}, \end{align*} where the last inclusion holds because $\mathfrak{a}$ is an ideal containing each $x_k$. Hence $\mathfrak{a} = (x_1, \ldots, x_m)$ is finitely generated. So $\mathcal{O}_L$ is Noetherian. [guided] Axiom (ii) — Noetherianity — is equivalent to the ascending chain condition on ideals, or equivalently to: every ideal is finitely generated. We verify the latter. **Strategy.** Since $\mathcal{O}_L \cong \mathbb{Z}^n$ as abelian groups, any ideal $\mathfrak{a}$ is, via this isomorphism, a subgroup of the free abelian group $\mathbb{Z}^n$. Subgroups of $\mathbb{Z}^n$ are well-controlled: they are themselves free abelian of rank at most $n$. In particular, they are finitely generated as abelian groups. **The subgroup fact.** Let $H \leq \mathbb{Z}^n$ be any subgroup. We claim $H$ is finitely generated as an abelian group. One proof: by induction on $n$. For the base case $n = 0$, $\mathbb{Z}^0 = 0$ so $H = 0$ is generated by the empty set. For the inductive step, consider the projection $\pi: \mathbb{Z}^n \to \mathbb{Z}$ onto the last coordinate. The image $\pi(H) \leq \mathbb{Z}$ is a subgroup of $\mathbb{Z}$, hence of the form $d\mathbb{Z}$ for some $d \geq 0$. The kernel $\ker(\pi|_H) = H \cap \mathbb{Z}^{n-1}$ is a subgroup of $\mathbb{Z}^{n-1}$, finitely generated by inductive hypothesis. Picking a preimage in $H$ of the generator of $d\mathbb{Z}$ (if $d \neq 0$), together with the generators of $H \cap \mathbb{Z}^{n-1}$, gives a finite generating set for $H$. **Applying to our ideal.** Let $\mathfrak{a} \unlhd \mathcal{O}_L$. The isomorphism $\Phi: \mathbb{Z}^n \xrightarrow{\sim} \mathcal{O}_L$ from [Existence of an Integral Basis](/theorems/1581) carries $\Phi^{-1}(\mathfrak{a}) \leq \mathbb{Z}^n$, which is a subgroup. By the fact above, $\Phi^{-1}(\mathfrak{a})$ is finitely generated as an abelian group; transporting the generators via $\Phi$, $\mathfrak{a}$ is finitely generated as an abelian group. Say \begin{align*} \mathfrak{a} = \mathbb{Z} x_1 + \mathbb{Z} x_2 + \cdots + \mathbb{Z} x_m \end{align*} for some $x_1, \ldots, x_m \in \mathfrak{a}$. **Upgrading to ideal generation.** Any $\mathbb{Z}$-generating set is automatically an ideal generating set, because $\mathcal{O}_L$-combinations include $\mathbb{Z}$-combinations: \begin{align*} \mathbb{Z} x_1 + \cdots + \mathbb{Z} x_m \subseteq \mathcal{O}_L x_1 + \cdots + \mathcal{O}_L x_m \subseteq \mathfrak{a}. \end{align*} The left inclusion is $\mathbb{Z} \subseteq \mathcal{O}_L$; the right inclusion is $x_k \in \mathfrak{a}$ combined with $\mathfrak{a}$ being an ideal (so $\mathcal{O}_L x_k \subseteq \mathfrak{a}$). The leftmost expression equals $\mathfrak{a}$ by assumption, and the rightmost is $\mathfrak{a}$ by construction, so all three sets are equal, giving \begin{align*} \mathfrak{a} = \mathcal{O}_L x_1 + \cdots + \mathcal{O}_L x_m = (x_1, \ldots, x_m). \end{align*} So every ideal of $\mathcal{O}_L$ has a finite generating set (of size $\leq n$, in fact, though we do not need this refinement). Hence $\mathcal{O}_L$ is Noetherian. [/guided] [/step] [step:Verify (iii): $\mathcal{O}_L$ is integrally closed in $L$] By [Fraction Field of the Ring of Integers](/theorems/1571), $\operatorname{Frac}(\mathcal{O}_L) = L$. Let $x \in L$ be integral over $\mathcal{O}_L$, meaning there exist $a_0, \ldots, a_{m-1} \in \mathcal{O}_L$ with \begin{align*} x^m + a_{m-1} x^{m-1} + \cdots + a_1 x + a_0 = 0. \tag{$\ast$} \end{align*} We show $x \in \mathcal{O}_L$. By definition of $\mathcal{O}_L$, every $a_k$ is integral over $\mathbb{Z}$. By [Transitivity of Integrality](/theorems/1568), applied with the tower $\mathbb{Z} \subseteq \mathcal{O}_L \subseteq L$: - Each $a_k$ is integral over $\mathbb{Z}$ (given). - $x$ is integral over $\mathcal{O}_L$ (given by $(\ast)$). - Therefore $x$ is integral over $\mathbb{Z}$. Since $x \in L$ and $x$ is integral over $\mathbb{Z}$, $x \in \mathcal{O}_L$ by the very definition of the ring of integers. [guided] Axiom (iii) requires that $\mathcal{O}_L$ equals its integral closure inside $L = \operatorname{Frac}(\mathcal{O}_L)$. Concretely: if $x \in L$ satisfies a monic polynomial with coefficients in $\mathcal{O}_L$, then $x \in \mathcal{O}_L$. **Hypothesis.** Suppose $x \in L$ and there exist $a_0, \ldots, a_{m-1} \in \mathcal{O}_L$ with \begin{align*} x^m + a_{m-1} x^{m-1} + \cdots + a_1 x + a_0 = 0. \end{align*} **Strategy: transitivity of integrality.** We have integrality at two levels: - Each $a_k \in \mathcal{O}_L$ is **integral over $\mathbb{Z}$** (this is the definition of $\mathcal{O}_L$ — elements of $L$ integral over $\mathbb{Z}$). - $x$ is **integral over $\mathcal{O}_L$** (from the displayed equation). We want to combine these into: $x$ is integral over $\mathbb{Z}$. This is exactly the content of [Transitivity of Integrality](/theorems/1568): if $A \subseteq B \subseteq C$ are commutative rings with $B$ integral over $A$ and $c \in C$ integral over $B$, then $c$ is integral over $A$. **Verifying hypotheses of Transitivity of Integrality.** We apply with $A = \mathbb{Z}$, $B = \mathcal{O}_L$, $C = L$: - $\mathbb{Z} \subseteq \mathcal{O}_L \subseteq L$: yes, these are ring inclusions. - $\mathcal{O}_L$ is integral over $\mathbb{Z}$: by definition, every element of $\mathcal{O}_L$ is integral over $\mathbb{Z}$. - $x \in L$ is integral over $\mathcal{O}_L$: from $(\ast)$. The theorem gives: $x$ is integral over $\mathbb{Z}$. **Conclusion.** $x \in L$ is integral over $\mathbb{Z}$. By definition $\mathcal{O}_L = \{y \in L : y \text{ integral over } \mathbb{Z}\}$, so $x \in \mathcal{O}_L$. This is the characteristic "closed under algebraic integrality within its fraction field" property — the integral closure of $\mathcal{O}_L$ in $L$ is $\mathcal{O}_L$ itself, making $\mathcal{O}_L$ integrally closed. [/guided] [/step] [step:Verify (iv): every nonzero prime ideal of $\mathcal{O}_L$ is maximal] Let $\mathfrak{p} \unlhd \mathcal{O}_L$ be a nonzero prime ideal. We show $\mathfrak{p}$ is maximal by showing the quotient $\mathcal{O}_L/\mathfrak{p}$ is a field. By [Nonzero Ideals Have Bounded Quotients](/theorems/1583), the quotient $\mathcal{O}_L/\mathfrak{p}$ is a finite set. By primality of $\mathfrak{p}$, the quotient $\mathcal{O}_L/\mathfrak{p}$ is an integral domain. We show: [claim:A finite integral domain is a field] Let $R$ be a finite integral domain with $|R| \geq 2$. Then $R$ is a field. [/claim] [proof] Let $x \in R$ with $x \neq 0$. We produce a multiplicative inverse. Consider the map \begin{align*} m_x: R &\to R \\ y &\mapsto x y. \end{align*} We verify $m_x$ is injective: if $m_x(y_1) = m_x(y_2)$ then $x(y_1 - y_2) = 0$, and since $R$ is an integral domain with $x \neq 0$, we deduce $y_1 - y_2 = 0$, i.e., $y_1 = y_2$. An injective map between finite sets of equal cardinality is automatically surjective (pigeonhole). Since $R$ is finite and $m_x: R \to R$ is injective, $m_x$ is surjective. In particular, there exists $y \in R$ with $m_x(y) = xy = 1$. This $y$ is a multiplicative inverse of $x$ in $R$. Since $x$ was an arbitrary nonzero element, every nonzero element of $R$ is a unit, so $R$ is a field. [/proof] Applying the claim to $R = \mathcal{O}_L/\mathfrak{p}$: the quotient is a finite integral domain, hence a field. Equivalence "$R/\mathfrak{p}$ is a field iff $\mathfrak{p}$ is maximal" (standard ring-theoretic fact) then gives $\mathfrak{p}$ maximal. Moreover we check $\mathcal{O}_L/\mathfrak{p}$ has more than one element: if $|\mathcal{O}_L/\mathfrak{p}| = 1$ then $\mathfrak{p} = \mathcal{O}_L$, but a prime ideal is by definition proper ($\mathfrak{p} \neq \mathcal{O}_L$), so $|\mathcal{O}_L/\mathfrak{p}| \geq 2$, and the claim applies. [guided] Axiom (iv) says: for every nonzero prime $\mathfrak{p} \unlhd \mathcal{O}_L$, $\mathfrak{p}$ is maximal. Equivalently (standard ring theory), the quotient $\mathcal{O}_L/\mathfrak{p}$ is a field. **The chain of implications.** 1. $\mathfrak{p}$ prime $\implies \mathcal{O}_L/\mathfrak{p}$ is an integral domain (by definition of prime ideal — an ideal $\mathfrak{p}$ is prime iff $R/\mathfrak{p}$ is a nonzero integral domain). 2. $\mathfrak{p}$ nonzero $\implies \mathcal{O}_L/\mathfrak{p}$ is finite (by [Nonzero Ideals Have Bounded Quotients](/theorems/1583); we spell this out below). 3. Finite integral domain $\implies$ field (proved in the claim below). 4. $\mathcal{O}_L/\mathfrak{p}$ field $\implies \mathfrak{p}$ maximal (standard ring theory: maximal ideals are exactly those whose quotient is a field). **Why [Nonzero Ideals Have Bounded Quotients](/theorems/1583) applies.** Its hypothesis is that $\mathfrak{a} \unlhd \mathcal{O}_L$ is a *nonzero* ideal; its conclusion is that $\mathcal{O}_L/\mathfrak{a}$ is finite. Our $\mathfrak{p}$ is nonzero by assumption, so directly $\mathcal{O}_L/\mathfrak{p}$ is finite. **The claim "finite integral domain is a field".** Let $R$ be a finite integral domain (with $|R| \geq 2$, which holds because $R$ nontrivial: $0 \neq 1$ in an integral domain by convention). We show every nonzero $x \in R$ has an inverse. Consider multiplication-by-$x$: $m_x: R \to R$, $y \mapsto xy$. This is a function from the finite set $R$ to itself. Two observations: - $m_x$ is injective: if $xy_1 = xy_2$ then $x(y_1 - y_2) = 0$; since $R$ is an integral domain and $x \neq 0$, we get $y_1 = y_2$. - An injective self-map of a finite set is a bijection (pigeonhole principle). By bijectivity, there exists $y \in R$ with $m_x(y) = 1$, i.e., $xy = 1$. So $x$ has an inverse in $R$. **Nontriviality of $\mathcal{O}_L/\mathfrak{p}$.** A prime ideal by definition is proper: $\mathfrak{p} \neq \mathcal{O}_L$. Hence $\mathcal{O}_L/\mathfrak{p}$ has more than one element, i.e., $|\mathcal{O}_L/\mathfrak{p}| \geq 2$. The claim applies. **Putting it together.** $\mathcal{O}_L/\mathfrak{p}$ is a finite integral domain of size $\geq 2$, hence a field. Therefore $\mathfrak{p}$ is maximal. **Why the nonzero assumption matters.** The zero ideal $(0) \unlhd \mathcal{O}_L$ is always prime (since $\mathcal{O}_L$ is an integral domain), but generally not maximal — its quotient is $\mathcal{O}_L$, which is not a field (for $n \geq 2$). The Dedekind domain axiom talks about *nonzero* primes. The result "nonzero primes are maximal" means $\dim \mathcal{O}_L = 1$ in Krull-dimensional terms: the only prime chain is $(0) \subsetneq \mathfrak{p}$, with no room to extend above $\mathfrak{p}$. [/guided] [/step] [step:Assemble (i)–(iv) to conclude] Combining: - (i) $\mathcal{O}_L$ is an integral domain (Step 2). - (ii) $\mathcal{O}_L$ is Noetherian (Step 3). - (iii) $\mathcal{O}_L$ is integrally closed in $L = \operatorname{Frac}(\mathcal{O}_L)$ (Step 4). - (iv) Every nonzero prime ideal of $\mathcal{O}_L$ is maximal (Step 5). By definition, $\mathcal{O}_L$ is a Dedekind domain. [guided] We have verified each of the four Dedekind domain axioms separately. Let us recap the structural flow: - **Integral domain** (i): inherited from $\mathcal{O}_L \subseteq L$, a field. - **Noetherian** (ii): because ideals, being subgroups of $\mathcal{O}_L \cong \mathbb{Z}^n$, are finitely generated abelian groups, hence a fortiori finitely generated ideals. The critical input was [Existence of an Integral Basis](/theorems/1581). - **Integrally closed** (iii): because integrality is transitive over $\mathbb{Z}$. The critical input was [Transitivity of Integrality](/theorems/1568), combined with the very definition of $\mathcal{O}_L$ as "the integral closure of $\mathbb{Z}$ in $L$". - **Nonzero primes maximal** (iv): because nonzero quotients are finite integral domains, hence fields. The critical input was [Nonzero Ideals Have Bounded Quotients](/theorems/1583). These four conditions together are the definition of a Dedekind domain. Consequence: the full structural theory of Dedekind domains applies to $\mathcal{O}_L$ — in particular, unique factorization of ideals into prime ideals (which can fail for elements, since $\mathcal{O}_L$ need not be a UFD, but always holds at the level of ideals), the group structure on fractional ideals, and the finite class group. These results form the backbone of algebraic number theory. [/guided] [/step]